Question

The potential energy between the atoms in a molecule is given by
$U\left(x\right)=\frac{a}{x^{12}}-\frac{b}{x^6}$

Where $a$ and $b$ are positive constants, and $x$ is the distance between the atoms. The atoms will be in equilibrium when,

• A. $x=0$
• B. $x={\left(\frac{a}{2b}\right)}^{1/6}$
• C. $x={\left(\frac{2a}{b}\right)}^{1/6}$
• D. $x={\left(\frac{11a}{5b}\right)}^{1/6}$

Answer 1

The correct option is C $x={\left(\frac{2a}{b}\right)}^{1/6}$

At equilibrium, potential energy has minimum value. So at equilibrium position,

$\frac{dU}{dx}=0$

$\frac{d}{dx}(\frac{a}{x^{12}}-\frac{b}{x^6})=0$

$(-12)a{x}^{-13}-(-6)b{x}^{-7}=0$

$\therefore x={\left(\frac{2a}{b}\right)}^{\frac{1}{6}}$

Hence, option $\left(C\right)$ is correct.

Why this question ? Tip: If potential energy is a function of $x$, then force on the particle is $F\left(x\right)=-\frac{dU}{dx}$ And, at equilibrium, $F=0$.